H2O+H3PO4+H3O++H2PO−4,pk1=2.15 H2O+H2PO−4+H3O++HPO2−4,pk2=7.20 p – InterviewSolution

InterviewSolution

1.	H2O+H3PO4+H3O++H2PO−4,pk1=2.15 H2O+H2PO−4+H3O++HPO2−4,pk2=7.20 pH of 0.01 M NaH2PO4 is:
Answer» $H_{2} O + H_{3} P O_{4} + H_{3} O^{+} + H_{2} P O_{4}^{-}, p^{k_{1}} = 2.15$ $H_{2} O + H_{2} P O_{4}^{-} + H_{3} O^{+} + H P O_{4}^{2 -}, p^{k_{2}} = 7.20$ $p^{H}$ of 0.01 M $N a H_{2} P O_{4}$ is:

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